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### IRWIN 9e 14_FE-2

Course: EE 2120, Fall 2008
School: LSU
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Basic Irwin, Engineering Circuit Analysis, 9/E 1 SOLUTION: The correct answer is c. 4 I ( s) 2 sI ( s ) 2 I ( s ) s 4 V s ( s ) 2 s 2 I ( s ) s V0 ( s ) 2 I ( s ) V (s) (s) I 0 2 4 V ( s) Vs ( s ) 2 s 2 0 s 2 V0 ( s ) s 2 Vs ( s ) s s 2 Vs ( s ) Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.FE-2

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Basic Irwin, Engineering Circuit Analysis, 9/E 1 SOLUTION: The correct answer is c. 4 I ( s) 2 sI ( s ) 2 I ( s ) s 4 V s ( s ) 2 s 2 I ( s ) s V0 ( s ) 2 I ( s ) V (s) (s) I 0 2 4 V ( s) Vs ( s ) 2 s 2 0 s 2 V0 ( s ) s 2 Vs ( s ) s s 2 Vs ( s ) Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.FE-2
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LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a. s4 I s (s) I 0 ( s) s4 3 s I 0 (s) s( s 4) 2 I s (s) s 4s 3Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-3
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. KVL:4s 4 s 2 I (s) s s 1 2I (s) 4s 2 ( s 2 1)( s 2 2 s 4)Chapter 14: Application of the Laplace Transform To Circuit AnalysisProblem 14.FE-4
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. KVL: Vs ( s ) 4 I ( s) 2 sI ( s )Vs ( s ) (2 s 4) I ( s ) V0 ( s) 2 sI ( s) V (s) I (s) 0 2sV s ( s ) ( 2 s 4) V0 ( s ) 2sV0 ( s ) s Vs ( s ) s 2Chapter 14: Applicatio
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
bn = 0
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
(s)(s)-(s) =(s) =(s) =(s)(s)(s)(s)o (s)(s) =(s)(s)-(s)(s)(s)(s)(s)(s)(s)(s)(s)(s)-
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
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LSU - EE - 2120
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LSU - EE - 2120
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LSU - EE - 2120
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LSU - EE - 2120
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b. By observation, the waveform has odd symmetry. Therefore, an = 0 for all n due to odd symmetry.Chapter 15: Fourier Analysis TechniquesProblem 15.FE-1
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. By observation, the waveform has half-wave symmetry. Therefore, bn =0 for n even due to half-wave symmetry and bn is nonzero for n odd.Chapter 15: Fourier Analysis Techni
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a. j V0 (n ) 1 j Vs (n ) V0 (0) 0 j 2 30 V0 ( ) 1 j 2 0 V0 ( ) 8.5426.57V j 4 30 V0 (2 ) 1 j 4 2 0 V0 (2 ) 4.6314.04V j 6 30 V0 (3 ) 1 j 6 3 0 V0 (3 ) 3.149.46V j8 30 V0 (
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b.I (n ) I DC Vs 2 jn3.7720 10 A 2 1060 I1 2.34 2.05 A 2 j 3.77 445 I3 0.35 34.97 A 2 j11.31 i (t ) 10 2.34 cos(377t 2.05) 0.35 cos(1131t 34.97) A VI P VDC I DC n n cos(
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b.a0 a0 1 1 10 dt 4 2 dt 40 2241 2 10t 0 1 2t 4 2 4 4 1 1 a 0 (20) 2(4) (2)(2) 4 4 a 0 4VChapter 15: Fourier Analysis TechniquesProblem 15.FE-5
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120